Difference between revisions of "2018 AMC 12B Problems/Problem 13"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Coordinate Geometry)) |
MRENTHUSIASM (talk | contribs) (Added in Sol 2--we don't need to prove that G1G2G3G4 is a square.) |
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~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | ~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | ||
− | ==Solution 2 (Coordinate Geometry)== | + | ==Solution 2 (Similar Triangles)== |
+ | This solution refers to the diagram in Solution 1. | ||
+ | |||
+ | By SAS, we conclude that <math>\triangle G_1G_3P\sim\triangle M_1M_3P</math> and <math>\triangle G_2G_4P\sim\triangle M_2M_4P.</math> By the properties of centroids, the ratio of similitude for each pair of triangles is <math>\frac23.</math> | ||
+ | |||
+ | Note that quadrilateral <math>M_1M_2M_3M_4</math> is a square of diagonal-length <math>30,</math> so <math>\overline{M_1M_3}\perp\overline{M_2M_4}.</math> Since <math>\overline{G_1G_3}\parallel\overline{M_1M_3}</math> and <math>\overline{G_2G_4}\parallel\overline{M_2M_4}</math> by the Converse of the Corresponding Angles Postulate, we have <math>\overline{G_1G_3}\perp\overline{G_2G_4}.</math> | ||
+ | |||
+ | Therefore, the area of quadrilateral <math>G_1G_2G_3G_4</math> is <cmath>\frac12\cdot G_1G_3\cdot G_2G_4 = \frac12\cdot\left(\frac23\cdot M_1M_3\right)\cdot\left(\frac23\cdot M_2M_4\right)=\boxed{\textbf{(C) }200}.</cmath> | ||
+ | ~Funnybunny5246 ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Coordinate Geometry)== | ||
We put the diagram on a coordinate plane. The coordinates of the square are <math>(0,0),(30,0),(30,30),(0,30)</math> and the coordinates of point <math>P</math> are <math>(x,y).</math> By using the centroid formula, we find that the coordinates of the centroids are <math>\left(\frac{x}{3},10+\frac{y}{3}\right),\left(10+\frac{x}{3},\frac{y}{3}\right),\left(20+\frac{x}{3},10+\frac{y}{3}\right),</math> and <math>\left(10+\frac{x}{3},20+\frac{y}{3}\right).</math> Shifting the coordinates down by <math>\left(\frac x3,\frac y3\right)</math>does not change its area, and we ultimately get that the area is equal to the area covered by <math>(0,10),(10,0),(20,10),(10,20)</math> which has an area of <math>\boxed{\textbf{(C) }200}.</math> | We put the diagram on a coordinate plane. The coordinates of the square are <math>(0,0),(30,0),(30,30),(0,30)</math> and the coordinates of point <math>P</math> are <math>(x,y).</math> By using the centroid formula, we find that the coordinates of the centroids are <math>\left(\frac{x}{3},10+\frac{y}{3}\right),\left(10+\frac{x}{3},\frac{y}{3}\right),\left(20+\frac{x}{3},10+\frac{y}{3}\right),</math> and <math>\left(10+\frac{x}{3},20+\frac{y}{3}\right).</math> Shifting the coordinates down by <math>\left(\frac x3,\frac y3\right)</math>does not change its area, and we ultimately get that the area is equal to the area covered by <math>(0,10),(10,0),(20,10),(10,20)</math> which has an area of <math>\boxed{\textbf{(C) }200}.</math> | ||
Revision as of 12:45, 27 September 2021
Contents
Problem
Square has side length
. Point
lies inside the square so that
and
. The centroids of
,
,
, and
are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
Solution 1 (Similar Triangles)
As shown below, let be the midpoints of
respectively, and
be the centroids of
respectively.
By SAS, we conclude that
and
By the properties of centroids, the ratio of similitude for each pair of triangles is
Note that quadrilateral is a square of side-length
It follows that:
- Since
and
by the Converse of the Corresponding Angles Postulate, we have
- Since
and
by the ratio of similitude, we have
Together, quadrilateral is a square of side-length
so its area is
Remark
This solution shows that, if point is within square
then the shape and the area of quadrilateral
are independent of the location of
Let the brackets denote areas. More generally,
is always a square of area
On the other hand, the location of
is dependent on the location of
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM
Solution 2 (Similar Triangles)
This solution refers to the diagram in Solution 1.
By SAS, we conclude that and
By the properties of centroids, the ratio of similitude for each pair of triangles is
Note that quadrilateral is a square of diagonal-length
so
Since
and
by the Converse of the Corresponding Angles Postulate, we have
Therefore, the area of quadrilateral is
~Funnybunny5246 ~MRENTHUSIASM
Solution 3 (Coordinate Geometry)
We put the diagram on a coordinate plane. The coordinates of the square are and the coordinates of point
are
By using the centroid formula, we find that the coordinates of the centroids are
and
Shifting the coordinates down by
does not change its area, and we ultimately get that the area is equal to the area covered by
which has an area of
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1439
~ pi_is_3.14
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.